HAL Id: hal-01131955
https://hal.archives-ouvertes.fr/hal-01131955v3
Preprint submitted on 14 Mar 2019
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
A characterization of class groups via sets of lengths
Alfred Geroldinger, Wolfgang Schmid
To cite this version:
Alfred Geroldinger, Wolfgang Schmid. A characterization of class groups via sets of lengths. 2018.
�hal-01131955v3�
ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
Abstract. LetH be a Krull monoid with class groupGsuch that every class contains a prime divisor.
Then every nonunita∈H can be written as a finite product of irreducible elements. Ifa=u1·. . .·uk, with irreduciblesu1, . . . , uk∈H, thenkis called the length of the factorization and the setL(a) of all possiblekis the set of lengths ofa. It is well-known that the systemL(H) ={L(a)|a∈H}depends only on the class group G. We study the inverse question asking whether the systemL(H) is characteristic for the class group. LetH′be a further Krull monoid with class groupG′such that every class contains a prime divisor and suppose thatL(H) =L(H′). We show that, if one of the groupsGandG′is finite and has rank at most two, thenGandG′are isomorphic (apart from two well-known exceptions).
1. Introduction
LetH be a cancellative semigroup with unit element. If an elementa∈H can be written as a product of k irreducible elements, say a=u1·. . .·uk, then k is called the length of the factorization. The set L(a) of all possible factorization lengths is the set of lengths of a, andL(H) ={L(a)|a ∈H} is called the system of sets of lengths of H. Clearly, ifH is factorial, then|L(a)| = 1 for each a∈ H. Suppose there is somea∈H with|L(a)|>1, sayk, l∈L(a) withk < l. Then, for everym∈N, we observe that L(am)⊃ {km+ν(l−k)|ν∈[0, m]}which shows that sets of lengths can become arbitrarily large. Under mild conditions on the ideal theory ofH every nonunit ofH, has a factorization into irreducibles and all sets of lengths are finite.
Sets of lengths (together with parameters controlling their structure) are the most investigated invari- ants in factorization theory. They occur in settings ranging from numerical monoids, noetherian domains, monoids of ideals and of modules to maximal orders in central simple algebras (for recent progress see [4, 6, 12]). The focus of the present paper is on Krull monoids with finite class group such that every class contains a prime divisor. Rings of integers in algebraic number fields are such Krull monoids, and classical notions from algebraic number theory (dating back to the 19th century) state that the class group determines the arithmetic of the ring of integers. This idea has been formalized and justified. In the 1970s Narkiewicz posed the inverse question whether or not arithmetical phenomena (in other words, phenomena describing the non-uniqueness of factorizations) characterize the class group ([27, Problem 32; page 469]). Very quickly first affirmative answers were given by Halter-Koch, Kaczorowski, and Rush ([26, 23, 29]). Indeed, it is not too difficult to show that the system of sets of factorizations determines the class group ([15, Sections 7.1 and 7.2]).
These answers are not really satisfactory because the given characterizations are based on rather ab- stract arithmetical properties which play only little role in other parts of factorization theory. Since, on the other hand, sets of lengths are of central interest in factorization theory, it is natural to ask whether their structure is rich enough to do characterizations.
LetH be a commutative Krull monoid with finite class groupGand suppose that every class contains a prime divisor (recall that an integral domain is a Krull domain if and only if its monoid of nonzero elements is a Krull monoid). It is classical thatH is factorial if and only if |G|= 1, and by a result due
2010Mathematics Subject Classification. 11B30, 11R27, 13A05, 13F05, 20M13.
Key words and phrases. Krull monoids, class groups, arithmetical characterizations, sets of lengths, zero-sum sequences.
This work was supported by the Austrian Science Fund FWF, Project Number P26036-N26, by the Austrian-French Amad´ee Program FR03/2012, and by the ANR Project Caesar, Project Number ANR-12-BS01-0011.
1
to Carlitz in 1960 we know that all sets of lengths are singletons (i.e., |L|= 1 for all L ∈ L(H)) if and only if|G| ≤2. Let us suppose now that |G| ≥3. Then the monoid B(G) of zero-sum sequences over Gis again a Krull monoid with class group isomorphic to G, every class contains a prime divisor, and L(H) =L B(G)
(as usual, we setL(G) =L B(G)
. The Characterization Problem can be formulated as follows ([15, Section 7.3], [17, page 42], [33]).
Given two finite abelian groups GandG′ such that L(G) =L(G′). Does it follow thatG∼=G′? The system of sets of lengths L(G) for finite abelian groups is studied with methods from Additive Combinatorics. Zero-sum theoretical invariants, such as the Davenport constant, play a central role.
Recall that, although the precise value of the Davenport constant is well-known for p-groups and for groups of rank at most two (see Proposition 2.2), its precise value is unknown in general (even for groups of the form G=Cn3). Thus it is not surprising that all answers to the Characterization Problem so far have been restricted to very special groups including cyclic groups, elementary 2-groups, and groups of the formCn⊕Cn ([7, 32]). Apart from two well-known (trivial) pairings, the answer is always positive.
Starting fromCn⊕Cn, Zhong studied the Characterization Problem for groups of the formCnrin a series of papers ([21, 38, 37]). The goal of the present paper is to settle the Characterization Problem for groups of rank at most two. Here is our main result.
Theorem 1.1. Let Gbe an abelian group such that L(G) =L(Cn1⊕Cn2)wheren1, n2∈Nwith n1|n2
andn1+n2>4. ThenG∼=Cn1⊕Cn2.
The difficulty of the Characterization Problem stems from the fact that most sets of lengths over any finite abelian group are arithmetical progressions with difference 1 (see Proposition 3.1.3, or [15, Theorem 9.4.11]). IfGandG′ are finite abelian groups withG⊂G′, then clearlyL(G)⊂ L(G′). Thus, in order to characterize a groupG, we first have to find distinctive sets of lengths forG(i.e., sets of lengths which do occur inL(G), but in no other or only in a small number of further groups), and second we will have to show that certain sets are not sets of lengths inL(G). These distinctive sets of lengths for rank two groups are identified in Proposition 5.5 which is the core of our whole approach, and Proposition 4.1 provides sets which do not occur as sets of lengths for rank two groups. After gathering some background material in Section 2, we summarize key results on the structure of sets of lengths in Propositions 3.1. 3.2, and 3.3. Furthermore, we provide some explicit constructions which will turn out to be crucial (Propositions 3.4 – 3.7). After that, we are well-prepared for the main parts given in Sections 4 and 5.
2. The arithmetic of Krull monoids: Background
We gather the required tools from the algebraic and arithmetic theory of Krull monoids. Our notation and terminology are consistent with the monographs [15, 17, 22]. LetNdenote the set of positive integers, P⊂Nthe set of prime numbers and put N0 =N∪ {0}. For real numbersa, b∈R, we set [a, b] ={x∈ Z|a≤x≤b}. LetA, B ⊂Z be subsets of the integers. We denote byA+B ={a+b |a∈A, b∈B}
theirsumset, and by ∆(A) theset of(successive)distancesofA(that is,d∈∆(A) if and only ifd=b−a witha, b∈Adistinct and [a, b]∩A={a, b}). Fork∈N, we denote byk·A={ka|a∈A} thedilation ofAbyk. IfA⊂N, then theelasticityofAis defined as
ρ(A) = supnm
n |m, n∈Ao
= supA
minA ∈Q≥1∪ {∞}and we set ρ({0}) = 1.
Monoids and Factorizations. By amonoid, we mean a commutative semigroup with identity which satisfies the cancellation law (that is, ifa, b, care elements of the monoid withab=ac, thenb=cfollows).
The multiplicative semigroup of non-zero elements of an integral domain is a monoid. LetH be a monoid.
We denote byH× the set of invertible elements ofH, byA(H) the set of atoms (irreducible elements) of H, and by Hred =H/H× = {aH× | a∈H} the associated reduced monoid of H. A monoid F is free abelian, with basisP ⊂F, and we writeF =F(P) if everya∈F has a unique representation of the form
a= Y
p∈P
pvp(a), where vp(a)∈N0 with vp(a) = 0 for almost all p∈P ,
and we call
|a|F =|a|=X
p∈P
vp(a) the lengthofa . The monoid Z(H) = F A(Hred)
is called the factorization monoid of H, and the homomorphism π: Z(H) → Hred, defined by π(u) = u for each u ∈ A(Hred) is the factorization homomorphism of H. Fora∈H,
ZH(a) =Z(a) =π−1(aH×)⊂Z(H) is the set of factorizations of a , and LH(a) =L(a) =
|z|
z∈Z(a) ⊂N0 is the set of lengths ofa .
Thus H is factorial if and only if Hred is free abelian (equivalently, |Z(a)| = 1 for all a ∈ H). The monoidH is calledatomicifZ(a)6=∅for alla∈H (equivalently, every nonunit can be written as a finite product of irreducible elements). For the remainder of this work, we suppose thatH is atomic. Note that, L(a) ={0} if and only ifa∈H×, andL(a) ={1}if and only ifa∈ A(H). We denote by
L(H) ={L(a)|a∈H} thesystem of sets of lengthsofH , and by
∆(H) = [
L∈L(H)
∆(L) ⊂N theset of distancesofH . Fork∈N, we setρk(H) =kifH=H×, and
ρk(H) = sup{supL|L∈ L(H), k∈L} ∈N∪ {∞}, if H 6=H×. Then
ρ(H) = sup{ρ(L)|L∈ L(H)}= lim
k→∞
ρk(H)
k ∈R≥1∪ {∞}
is theelasticityofH. The monoidH is said to be
• half-factorial if ∆(H) =∅. IfH is not half-factorial, then min ∆(H) = gcd ∆(H).
• decomposableif there exist submonoidsH1, H2withHi6⊂H×fori∈[1,2] such thatH =H1×H2
(otherwise H is called indecomposable).
For a free abelian monoidF(P), we introduce a distance functiond:F(P)× F(P)→N0, by setting d(a, b) = maxn
a gcd(a, b)
,
b gcd(a, b)
o∈N0 for a, b∈ F(P),
and we note thatd(a, b) = 0 if and only ifa=b. For a subset Ω⊂ F(P), we define thecatenary degree c(Ω) as the smallest N ∈ N0∪ {∞} with the following property: for eacha, b ∈ Ω, there are elements a0, . . . ak ∈Ω such thata=a0, ak =b, and d(ai−1, ai)≤N for all i∈[1, k]. Note that c(Ω) = 0 if and only if |Ω| ≤1. For an elementa∈ H, we call cH(a) =c(a) =c(ZH(a)) the catenary degree ofa, and c(H) = sup{c(a)|a∈H} ∈N0∪ {∞}is thecatenary degreeofH. The monoidH is factorial if and only ifc(H) = 0, and if H is not factorial, then 2 + sup ∆(H)≤c(H) ([15, Theorem 1.6.3]).
Krull monoids. A monoidH isKrullif it is completely integrally closed and satisfies the ascending chain condition on divisorial ideals. An integral domain R is a Krull domain if and only if its multiplicative monoidR\ {0}is a Krull monoid, and this generalizes to Marot rings ([16]). The theory of Krull monoids is presented in detail in [25, 15], and for a survey we refer to [12].
Much of the arithmetic of a Krull monoid can be seen in an associated monoid of zero-sum sequences.
This is a Krull monoid again which can be studied with methods from Additive Combinatorics. To introduce the necessary concepts, let G be an additively written abelian group, G0 ⊂ Ga subset, and letF(G0) be the free abelian monoid with basisG0. In Additive Combinatorics, the elements ofF(G0) are called sequences over G0. If S = g1·. . .·gl ∈ F(G0), where l ∈ N0 and g1, . . . , gl ∈ G0, then σ(S) =g1+. . .+gl is called the sum ofS, and the monoid
B(G0) ={S∈ F(G0)|σ(S) = 0} ⊂ F(G0)
is called the monoid of zero-sum sequences over G0 (these objects are also referred to in the literature as block monoids). The embedding B(G0) ֒→ F(G0) is a divisor homomorphism andB(G0) is a Krull monoid. The monoidB(G) is factorial if and only if|G| ≤2. If|G| 6= 2, thenB(G) is a Krull monoid with class group isomorphic toGand every class contains precisely one prime divisor. For every arithmetical invariant ∗(H) defined for a monoidH, it is usual to write∗(G0) instead of∗(B(G0)). In particular, we setA(G0) =A(B(G0)) andL(G0) =L(B(G0)). Similarly, arithmetical properties ofB(G0) are attributed toG0. Thus,G0 is said to be
• (in)decomposable ifB(G0) is (in)decomposable,
• (non-)half-factorial ifB(G0) is (non-)half-factorial.
Proposition 2.1. Let H be a Krull monoid with class groupG, and suppose that each class contains a prime divisor. Then there is a transfer homomorphism β:H → B(G)such that the following hold.
1. LH(a) =LB(G) β(a)
for eacha∈H andL(H) =L(G).
2. If |G| ≥3, then c(H) =c B(G) .
Proof. See [15, Section 3.4].
The above result generalizes totransfer Krull monoidsH over abelian groupsG, which also satisfy the relationshipL(H) =L(G). Hence all characterization results, such as Theorem 1.1, also apply to them ([12]).
Zero-Sum Theory. Let Gbe an additive abelian group, G0 ⊂G a subset, andG•0 =G0\ {0}. Then [G0]⊂Gdenotes the subsemigroup andhG0i ⊂Gthe subgroup generated byG0. For a sequence
S=g1·. . .·gl= Y
g∈G0
gvg(S)∈ F(G0),
we set ϕ(S) = ϕ(g1)·. . .·ϕ(gl) for any homomorphism ϕ: G→ G′, and in particular, we have −S = (−g1)·. . .·(−gl). We call
supp(S) ={g∈G|vg(S)>0} ⊂Gthe support of S , vg(S) themultiplicityof gin S ,
|S|=l=X
g∈G
vg(S)∈N0 the length of S , and k(S) =X
g∈G
1
ord(g) ∈Qthe cross number of S . Moreover, Σ(S) = n
P
i∈Igi | ∅ 6=I ⊂ [1, l]o
is the set of subsequence sums of S. The sequence S is said to be
• zero-sum free if 0∈/ Σ(S),
• a zero-sum sequence if σ(S) = 0,
• aminimal zero-sum sequence if it is a nontrivial zero-sum sequence and every proper subsequence is zero-sum free.
Both Davenport constants, namely
• the (small) Davenport constant d(G0) = sup
|S|
S∈ F(G0) is zero-sum free ∈N0∪ {∞}and
• the (large)Davenport constant D(G0) = sup
|U|
U ∈ A(G0) ∈N0∪ {∞}
are classical invariants in zero-sum theory. For n ∈ N, let Cn denote a cyclic group with n elements.
Suppose thatGis finite. A tuple (ei)i∈I is called abasisofGif all elements are nonzero andG=⊕i∈Iheii.
Forp∈ P, letrp(G) denote the p-rank of G, r(G) = max{rp(G)| p∈P} denote the rank of G, and let r∗(G) =P
p∈Prp(G) be thetotal rankofG. If|G|>1, thenG∼=Cn1⊕. . .⊕Cnr, and we set d∗(G) = Pr
i=1(ni−1),wherer, n1, . . . , nr∈Nwith 1< n1|. . .|nr,r=r(G), andnr= exp(G) is the exponent of G. If |G|= 1, thenr(G) =r∗(G) = 0, exp(G) = 1, andd∗(G) = 0. We will use the following well-known results (see [15, Chapter 5]).
Proposition 2.2. Let Gbe a finite abelian group. Then 1 +d∗(G)≤1 +d(G) =D(G)≤ |G|. If Gis a p-group or r(G)≤2, thend(G) =d∗(G).
We will make substantial use of the following result (see [15, Theorem 6.4.7] and [19, Theorem 1.1]).
Proposition 2.3. Let H be a Krull monoid with finite class group G where |G| ≥ 3 and every class contains a prime divisor. Thenc(H)∈[3,D(G)], and we have
1. c(H) =D(G) if and only ifGis either cyclic or an elementary 2-group.
2. c(H) =D(G)−1if and only ifGis isomorphic either toC2r−1⊕C4 for somer≥2or to C2⊕C2n
for somen≥2.
LetA∈ B(G0) andd= min{|U| |U ∈ A(G0)}. IfA=BCwithB, C ∈ B(G0), then L(B) +L(C)⊂L(A).
IfA=U1·. . .·Uk =V1·. . .·Vl withU1, . . . , Uk, V1, . . . , Vl∈ A(G0) andk < l, then ld≤
l
X
ν=1
|Vν|=|A|=
k
X
ν=1
|Uν| ≤kD(G0), whence |A|
D(G0) ≤minL(A)≤maxL(A)≤ |A|
d .
We need the concept of relative block monoids (as introduced by Halter-Koch in [24], and recently studied by Baeth et al. in [3]). LetGbe an abelian group. For a subgroupK⊂Glet
BK(G) ={S∈ F(G)|σ(S)∈K} ⊂ F(G), and letDK(G) denote the smallestl∈N∪ {∞}with the following property:
• Every sequenceS∈ F(G) of length|S| ≥l has a subsequenceT withσ(T)∈K.
Clearly,BK(G)⊂ F(G) is a submonoid with
B(G) =B{0}(G)⊂ BK(G)⊂ BG(G) =F(G)
andD{0}(G) =D(G). The following result is well-known ([3, Theorem 2.2]). Since there seems to be no proof in the literature and we make substantial use of it, we provide the simple arguments.
Proposition 2.4. Let Gbe an abelian group andK⊂Ga subgroup.
1. BK(G) is a Krull monoid. If|G|= 2and K={0}, then BK(G) =B(G)is factorial. In all other cases the embedding BK(G)֒→ F(G) is a divisor theory with class group isomorphic to G/K and every class contains precisely |K|prime divisors.
2. The monoid homomorphism θ:BK(G)→ B(G/K), defined byθ(g1·. . .·gl) = (g1+K)·. . .·(gl+K) is a transfer homomorphism. If |G/K| ≥3, thenc BK(G)
=c B(G/K) . 3. DK(G) = sup{|U| |U is an atom ofBK(G)}=D(G/K).
Proof. 1. If|G|= 1, then B(G) =F(G) is factorial, the class group is trivial, and there is precisely one prime divisor. If |G| =|K| = 2, then BK(G) = F(G) is factorial, the class group is trivial, and there are precisely two prime divisors. If|G| = 2 and |K|= 1, then BK(G) =B(G) is factorial, and hence a Krull monoid with trivial class group. Suppose that|G| ≥3. Clearly, the embeddingBK(G)֒→ F(G) is a divisor homomorphism. To verify that it is a divisor theory, letg∈Gbe given. If ord(g) =n≥3, then g= gcd gn, g(−g)
. If ord(g) = 2, then there is an elementh∈G\ {0, g}andg= gcd g2, gh(g−h) . If ord(g) =∞, then g= gcd (−g)g, g(2g)(−3g)
.
The mapϕ:F(G)→G/K, defined byϕ(S) =σ(S)+Kfor everyS∈ F(G), is a monoid epimorphism.
If S, S′ ∈ F(G), then σ(S) +K = σ(S′) +K if and only if S′ ∈ [S] = Sq BK(G)
. Thus ϕ induces a group isomorphism Φ : q(F(G))/q(BK(G)) → G/K, defined by Φ([S]) = σ(S) +K, and we have [S]∩G=σ(S) +K. Thus the class [S] contains precisely|K|prime divisors.
2. This follows from 1., from Proposition 2.1, and from the reference given there.
3. Since a sequenceS∈ F(G) is an atom ofBK(G) if and only ifS6= 1,σ(S)∈K andσ(T)∈/K for all proper subsequencesT of S, it follows that DK(G) = sup{|U| |U is an atom of BK(G)}. Since θ is a transfer homomorphism, we getθ A(BK(G))
=A(G/K) andθ−1 A(G/K)
=A(BK(G)). Therefore
|U|=|θ(U)|for allU∈ BK(G), and it follows that
sup{|U| |U ∈ A(BK(G))}= sup{|V| |V ∈ A(G/K)}=D(G/K).
3. Structural results onL(G) and first basic constructions
LetGbe an abelian group. IfGis infinite, then every finite subsetL⊂N≥2is contained inL(G) ([15, Theorem 7.4.1]). IfGis finite, then sets of lengths have a well-studied structure. In order to describe it, we recall the concept of an AAMP. Letd∈N, l, M ∈N0 and {0, d} ⊂ D ⊂[0, d]. A subsetL⊂Zis called analmost arithmetical multiprogression (AAMP for short) with difference d, period D, length land bound M, if
(3.1) L=y+ (L′∪L∗∪L′′) ⊂ y+D+dZ
where minL∗ = 0, L∗ is an interval of D+dZ (this means that L∗ is finite nonempty and L∗ = (D+dZ)∩[0,maxL∗]), l is maximal such that ld∈ L∗, L′ ⊂[−M,−1], L′′ ⊂maxL∗+ [1, M] and y∈Z. Theset of minimal distances ∆∗(G)⊂∆(G) is defined as
∆∗(G) ={min ∆(G0)|G0⊂Gwith ∆(G0)6=∅} ⊂∆(G).
Proposition 3.1(Structural results onL(G)). Let Gbe a finite abelian group with |G| ≥3.
1. There exists some M ∈ N0 such that every set of lengths L ∈ L(G) is an AAMP with some difference d∈∆∗(G) and boundM.
2. For every M ∈N0 and every finite nonempty set ∆∗ ⊂N, there is a finite abelian groupG∗ such that for every AAMP Lwith difference d∈∆∗ and boundM there is ayL ∈Nwith
y+L∈ L(G∗) for all y≥yL.
3. If A∈ B(G)such that supp(A)∪ {0} is a subgroup ofG, thenL(A) is an arithmetical progression with difference 1.
Proof. We refer to [15, Theorems 4.4.11 and 7.6.8]) and to [34].
Proposition 3.2(Structural results on ∆(G) and on ∆∗(G)).
LetG=Cn1⊕. . .⊕Cnr wherer, n1, . . . , nr∈Nwithr=r(G),1< n1|. . .|nr, and|G| ≥3.
1. ∆(G)is an interval with
1,max{exp(G)−2, k−1}
⊂∆(G)⊂
1,D(G)−2
where k=
r(G)
X
i=1
jni
2 k.
2. 1∈∆∗(G)⊂∆(G),[1,r(G)−1]⊂∆∗(G), andmax ∆∗(G) = max{exp(G)−2,r(G)−1}.
3. If Gis cyclic of order |G|=n≥4, thenmax ∆∗(G)\ {n−2}
=⌊n2⌋ −1.
Proof. We refer to [15, Section 6.8], to [18], and to [20].
Proposition 3.3(Results on ρk(G) and onρ(G)).
LetG be a finite abelian group with|G| ≥3, and let k∈N.
1. ρ(G) =D(G)/2 andρ2k(G) =kD(G).
2. 1 +kD(G)≤ρ2k+1(G)≤kD(G) +D(G)/2. IfGis cyclic, then equality holds on the left side.
Proof. We refer to [15, Chapter 6.3], [11, Theorem 5.3.1], and to [14] for recent progress.
In the next propositions we provide examples of sets of lengths over cyclic groups, over groups of rank two, and over groups of the form C2r−1⊕Cn with r, n ∈ N≥2. All examples will have difference d = max ∆∗(G) and period D with {0, d} ⊂ D ⊂ [0, d] and |D| = 3, and we write them down in a form used in Equation (3.1) in order to highlight their periods. It will be crucial for our approach (see Proposition 5.5) that the sets given in Proposition 3.4.2 do not occur over cyclic groups (Proposition 3.5). It is well-known that sets of lengths over cyclic groups and over elementary 2-groups have many features in common, and this carries over to rank two groups and groups of the form C2r−1⊕Cn (see Propositions 3.4.2, 3.6.2, and 5.5). For a set L ∈ L(G) there is a B ∈ B(G) such that L =L(B) and hencem+L=L(0mB)∈ L(G) for all m∈N0. Therefore the interesting sets of lengthsL ∈ L(G) are
those which do not stem from such a shift. These are those setsL∈ L(G) with−m+L /∈ L(G) for every m∈N.
Proposition 3.4. Let G=Cn1⊕Cn2 wheren1, n2∈Nwith2< n1|n2, and let d∈[3, n1].
1. For each k∈N, we have
(2k+ 2) +{0, d−2, n2−2}+{ν(n2−2)|ν∈[0, k−1]} ∪ {(kn+ 2) + (d−2)}= (2k+ 2) +{0, d−2}+{ν(n2−2)|ν∈[0, k]} ∈ L(G).
2. For each k∈N, we have
(2k+ 3) +{0, n1−2, n2−2}+{ν(n2−2)|ν∈[0, k]}
∪n
(kn2+ 3) + (n1−2) + (n2−2)o
∈ L(G). Proof. Let (e1, e2) be a basis of G with ord(ei) =ni fori ∈[1,2], and let k∈ N. For i ∈[1,2], we set Ui=enii andVi = (−ei)ei. Then
(−Ui)kUik= (−Ui)k−νUik−νViνni for all ν ∈[0, k], and hence
L (−Ui)kUik
= 2k+{ν(ni−2)|ν ∈[0, k]}.
1. We seth= (d−1)e1, W1= (−e1)d−1h, andW2=en11−(d−1)h. ThenZ(U1W1) ={U1W1, V1d−1W2} andL(U1W1) ={2, d}. Therefore
L (−U2)kU2kU1W1
=L (−U2)kU2k
+L U1W1
={2k+ν(n2−2)|ν∈[0, k]}+{2, d}
= (2k+ 2) +{ν(n2−2)|ν ∈[0, k]}+{0, d−2}. 2. We define
W1=en11−1en22−1(e1+e2), W2= (−e1)en22−1(e1+e2), W3=en11−1(−e2)(e1+e2), W4= (−e1)(−e2)(e1+e2), and Bk =W1(−U1)(−U2)U2k(−U2)k.
Then any factorization ofBk is divisible by precisely one ofW1, . . . , W4, and we obtain that Bk =W1(−U1)(−U2)U2k(−U2)k=W2V1n1−1(−U2)U2k(−U2)k
=W3(−U1)V2n2−1U2k(−U2)k =W4V1n1−1V2n2−1U2k(−U2)k. Thus it follows that
L(Bk) ={3, n1+ 1, n2+ 1, n1+n2−1}+L U2k(−U2)k
= (2k+ 3) +{ν(n2−2)|ν∈[0, k]} ∪
(2k+ 3) + (n1−2) +{ν(n2−2)|ν∈[0, k]} ∪ (2k+ 3) + (n2−2) +{ν(n2−2)|ν∈[0, k]} ∪
(2k+ 3) + (n1−2) + (n2−2) +{ν(n2−2)|ν ∈[0, k]}. Thus maxL(Bk) = (kn2+ 3) + (n1−2) + (n2−2) and
L(Bk) =
(2k+ 3) +{0, n1−2, n2−2}+{ν(n2−2)|ν ∈[0, k]}
∪ {maxL(Bk)}. Proposition 3.5. Let Gbe a cyclic group of order |G|=n≥4, and let d∈[3, n−1].
1. For each k∈N0, we have
(2k+ 2) +{0, d−2}+{ν(n−2)|ν∈[0, k]} ∈ L(G).
2. For each k∈N0, we set Lk =
(2k+ 3) +{0, d−2, n−2}+{ν(n−2)|ν∈[0, k]}
∪n
(kn+ 3) + (d−2) + (n−2)o . Then for each k∈N0 and each m∈N0, we have−m+Lk ∈ L(G)./
Proof. Letk∈N0.
1. Letg∈Gwith ord(g) =n,U =gn,V = (−g)g,W1= (d−1)g
(−g)d−1,W2= (d−1)g
gn−(d−1), andBk = (−U)Uk
U W1. ThenZ(U W1) ={U W1, W2Vd−1}and L(U W1) ={2, d}. Since every factor- ization ofBk is divisible either byW1 or byW2, it follows that
L(Bk) =L (−U)kUk
+L U W1
={2k+ν(n−2)|ν∈[0, k]}+{2, d}
= (2k+ 2) +{ν(n−2)|ν∈[0, k]}+{0, d−2}.
2. Note that maxLk= (kn+ 3) + (d−2) + (n−2) = (k+ 1)n+ (d−1). Assume to the contrary that there is aBk∈ B(G) such thatL(Bk) =Lk. Then minL(Bk) = 2k+ 3 and, by Proposition 3.3,
(k+ 1)n+ (d−1) = maxL(Bk)≤ρ2k+3(G) = (k+ 1)n+ 1,
a contradiction. Ifm∈N0andBm,k∈ B(G) such thatL(Bm,k) =−m+Lk, thenL(0mBm,k) =Lk∈ L(G).
Thus−m+Lk∈ L(G) for any/ m∈N0.
Proposition 3.6. Let G=C2r−1⊕Cn where r, n∈N≥2 andnis even.
1. For each k∈N0, we have
Lk = (2k+ 2) +{0, n−2, n+r−3}+{ν(n−2)|ν ∈[0, k]} ∈ L(G) yet Lk∈ L(C/ n). 2. For each k∈N0, we have
(2k+ 3) +{0, r−1, n−2}+{ν(n−2)|ν∈[0, k]}
∪n
(kn+ 3) + (r−1) + (n−2)o
∈ L(G). Proof. Letk∈N0, (e1, . . . , er−1, er) be a basis ofGwith ord(e1) =. . .= ord(er−1) = 2 and ord(er) =n.
We sete0=e1+. . .+er−1,Ui=eord(ei i) for eachi∈[1, r],U0= (e0+er)(e0−er),Vr= (−er)er, V =e1·. . .·er−1(e0+er)(−er), and W =e1·. . .·er−1(e0+er)en−1r .
1. Obviously,L (−W)W
={2, n, n+r−1}and L (−W)W(−Ur)kUrk
=L (−W)W
+L (−Ur)kUrk
={2, n, n+r−1}+{2k+ν(n−2)|ν∈[0, k]}
= (2k+ 2) +{0, n−2, n+r−3}+{ν(n−2)|ν ∈[0, k]}
Since minLk = 2k+ 2, maxLk = (k+ 1)n+r−1, and ρ2k+2(Cn) = (k+ 1)nby Proposition 3.3, r≥2 implies thatLk∈ L(C/ n).
2. LetLk denote the set in the statement. We define
Bk =U0U1·. . .·Ur−1(−Ur)k+1Urk+1
and assert thatL(Bk) =Lk. Letz be a factorization ofBk. We distinguish two cases.
CASE 1: U1|z.
ThenU0U1·. . . Ur−1|zwhich implies thatz=U0U1·. . .·Ur−1 (−Ur)Urk+1−ν
Vrνnfor someν∈[0, k+1]
and hence|z| ∈r+ (2k+ 2) +{ν(n−2)|ν∈[0, k+ 1]}.
CASE 2: U1∤z.
Then either V|z or W|z. If V |z, then z = (−V)V Vrn−1 (−Ur)Urk−ν
Vrνn for some ν ∈ [0, k] and hence|z| ∈(n+ 1) + 2k+{ν(n−2)|ν∈[0, k]}. IfW|z, thenz= (−W)W Vr (−Ur)Urk−ν
Vrνnfor some ν∈[0, k] and hence|z| ∈3 + 2k+{ν(n−2)|ν ∈[0, k]}. Putting all together the assertion follows.
Proposition 3.7. Let G be a finite abelian group,g ∈G withord(g) =n≥5, andB ∈ B(G) such that (−g)g2n
|B. Suppose L(B)is anAAMP with period {0, d, n−2}for somed∈[1, n−3]\ {(n−2)/2}.
1. If S∈ A Bhgi(G)
withS|B, thenσ(S)∈ {0, g,−g,(d+ 1)g,−(d+ 1)g}.
2. If S1, S2∈ A Bhgi(G)
with S1S2|B, thenσ(Si)∈ {0, g,−g}for at least onei∈[1,2].
Proof. By definition, there is ay∈Z such thatL(B)⊂y+{0, d, n−2}+ (n−2)Z. We setU =gn and V = (−g)g.
1. LetS ∈ A Bhgi(G)
withS|B and setσ(S) =kg withk∈[0, n−1]. If k∈ {0,1, n−1}, then we are done. Suppose thatk∈[2, n−2]. SinceS is an atom inBhgi(G), it follows thatW1=S(−g)k∈ A(G) and W1′ =Sgn−k ∈ A(G). We consider a factorization z ∈ Z(B) with U W1|z, say z = U W1y. Then z′ = W1′Vky is a factorization of B of length |z′| = |z|+k−1. Since L(B) is an AAMP with period {0, d, n−2}for somed∈[1, n−3]\ {(n−2)/2} it follows thatk−1∈ {d, n−2−d}.
2. LetS1, S2∈ A Bhei(G)
withS1S2|B, and assume to the contrary σ(Si) =kiewith ki ∈[2, n−2]
for eachi∈[1,2]. As in 1. it follows that
W1=S1(−g)k1, W1′ =S1gn−k1, W2=S2(−g)k2, and W2′=S2gn−k2
are in A(G). We consider a factorization z ∈ Z(B) with U W1U W2|z, say z = U W1U W2y. Then z1 = W1′Vk1−1U W2y ∈ Z(B) with |z1| = |z|+k1−1 and hence k1−1 ∈ {d, n−2−d}. Similarly, z2=U W1W2′Vk2−1y∈Z(B), hencek2−1∈ {d, n−2−d}, and furthermore it follows thatk1=k2. Now z3=W1′Vk1−1W2′Vk2−1y∈Z(B) is a factorization of length|z3|=|z|+k1+k2−2. Thus, ifk1−1 =d, then 2d∈ {n−2, n−2+d}, a contradiction, and ifk1−1 =n−2−d, then 2(n−2−d)∈ {n−2, n−2+(n−2−d)},
a contradiction.
4. A set of lengths not contained in L(Cn1⊕Cn2) The aim of this section is to prove the following proposition.
Proposition 4.1. Let G=Cn1⊕Cn2 wheren1, n2∈Nwithn1|n2 and6≤n1< n2. Then {2, n2, n1+n2−2}∈ L(G)./
LetG=Cn1⊕Cn2 wheren1, n2∈Nwithn1|n2. If 6≤n1< n2does not hold, then{2, n2, n1+n2−2}
may or may not be a set of lengths (e.g., if 2 = n1 ≤ n2, then {2, n2} ∈ L(G)). By Proposition 3.6, {2, n2, n1+n2−2} ∈ L(C2n1−2⊕Cn2), whence Proposition 4.1 implies thatL(Cn1⊕Cn2)6=L(C2n1−2⊕Cn2).
Its proof is based on the characterization of all minimal zero-sum sequences of maximal length over groups of rank two. This characterization is due to Gao, Grynkiewicz, Reiher, and the present authors ([8, 9, 35, 28]). We repeat the formulation given in [5, Theorem 3.1] and then derive a corollary.
Lemma 4.2. Let G = Cn1⊕Cn2 where n1, n2 ∈ N with 1 < n1|n2. A sequence U over G of length D(G) =n1+n2−1is a minimal zero-sum sequence if and only if it has one of the following two forms:
•
U =eord(ej j)−1
ord(ei)
Y
ν=1
(xνej+ei), where
(a) {i, j}={1,2}and(e1, e2)is a basis of Gwithord(e1) =n1 andord(e2) =n2, (b) x1, . . . , xord(ei)∈[0,ord(ej)−1]andx1+. . .+xord(ei)≡1 mod ord(ej).
In this case, we say that U is of type I with respect to the basis (ei, ej);
•
U = (e1+ye2)sn1−1en22−sn1+ǫ
n1−ǫ
Y
ν=1
(−xνe1+ (−xνy+ 1)e2), where (a) (e1, e2)is a basis ofGwith ord(e1) =n1 andord(e2) =n2,
(b) y∈[0, n2−1], ǫ∈[1, n1−1], and s∈[1, n2/n1−1],
(c) x1, . . . , xn1−ǫ∈[1, n1−1]with x1+. . .+xn1−ǫ=n1−1, (d) n1ye26= 0, and
(e) eithers= 1 orn1ye2=n1e2.
In this case, we say that U is of type II with respect to the basis (e1, e2).
We record some observations on this result. Ifn1 =n2, then sequences of type II do not exist as the conditionn1ye26= 0 cannot hold. Assume that n16=n2. There are examples of sequences that are both of type I and of type II. However, such sequences are of a rather special form.
If a sequenceU is of type I with j = 2, then it contains an element with multiplicity n2−1. Thus, U is of type II only when s= 1 andǫ=n1−1 and consequently x1 =n1−1, that is, U =en22−1(e1+ ye2)n1−1(e1+ (−(n1−1)y+ 1)e2) withy∈[0, n2−1] andn1ye26= 0. Such a sequence is indeed of type I.
If a sequenceU is of type I withj= 1, then it contains an element of ordern1with multiplicityn1−1.
Since the order ofe1+ye2 cannot ben1, asn1ye26= 0, and the order ofe2 is notn1 either, this is only possible whenǫ= 1 and consequentlyx1=· · ·=xn1−1= 1, that is,U = (e1+ye2)sn1−1en22−sn1+1(−e1+ (−y + 1)e2)n1−1. If n1ye2 = n1e2, then indeed e′1 = −e1+ (−y + 1)e2 is an element of order n1, we get that (e′1, e2) is a basis of G and U is of type I with respect to the basis (e′1, e2), indeed, U = e′n11−1((n1−1)e′1+e2)sn1−1en22−sn1+1 for somes∈[1, n2/n1−1].
Corollary 4.3. Let G=Cn1⊕Cn2 where n1, n2 ∈N with n1|n2 and 6≤n1 < n2, and let U ∈ A(G) with|U|=D(G) =n1+n2−1.
1. If h(U) = n2−1, then U is of type I with respect to a basis (e1, e2) with ord(e1) = n1 and ord(e2) =n2, that is
U =eord(e2 2)−1
ord(e1)
Y
ν=1
(xνe2+e1) wherex1, . . . , xn1∈[0, n2−1]withx1+. . .+xn1 ≡1 modn2. 2. If h(U) =n2−2, then
U = (e1+ye2)n1−1en22−2 −xe1+ (−xy+ 1)e2
−(n1−1−x)e1+ (−(n1−1−x)y+ 1)e2 , where(e1, e2)is a basis with ord(e1) =n1,ord(e2) =n2,y∈[0, n2−1], andx∈[1,(n1−1)/2].
3. If h(U) =n2−3, then
U = (e1+ye2)n1−1en22−3
3
Y
ν=1
(−xνe1+ (−xνy+ 1)e2),
where(e1, e2)is a basis withord(e1) =n1,ord(e2) =n2,y∈[0, n2−1], and x1, x2, x3∈[1, n1−1]
with x1+x2+x3≡n1−1 modn1 (ify 6= 0, thenx1+x2+x3=n1−1).
4. There is at most one element g ∈ Gwith vg(U)≥n2−3. In particular, if h(U)≥n2−3, then there is precisely one elementg∈Gwith vg(U) =h(U).
Proof. We use all notation as in Lemma 4.2.
1. IfU is of type II with respect to the basis (e1, e2), then as observed aboves= 1,ǫ=n1−1, and U = (e1+ye2)n1−1en22−1 e1+ ((−n1+ 1)y+ 1)e2
,
which shows that U is also of type I with respect to the basis (e1, e2). If U is of type I with respect to the basis (e2, e1) thenh(U) =n2−1 implies thatU is also of type I with respect to the basis (e1, e2).
2. Suppose thatU is of type I with respect to the basis (f2, f1). ThenU has the form U =f1n1−1(x1f1+f2)n2−2(x2f1+f2)(x3f1+f2).
ThusU has the asserted form withy= 0,e1=f1, and withe2=x1f1+f2. In this case we only have two summands the congruence condition modulon2, and hence we obtain an equality in the integers. Suppose
that U is of type II with respect to the basis (e1, e2). Then s = 1, ǫ =n1−2, and thus the assertion follows.
3. Suppose thatU is of type I with respect to the basis (f1, f2). ThenU has the form U =f1n1−1(x1f1+f2)n2−3(x2f1+f2)(x3f1+f2)(x4f1+f2).
ThusU has the asserted form withy= 0, e1=f1, and withe2=x1f1+f2. Suppose thatU is of type II with respect to the basis (e1, e2). Thens= 1,ǫ=n1−3, and thus the assertion follows.
4. Assume to the contrary that there are two distinct elements g1, g2 ∈Gwith vg1(U)≥n2−3 and vg2(U)≥n2−3. Then
(n2−3) + (n2−3)≤vg1(U) +vg2(U)≤ |U|=n1+n2−1,
which implies thatn2≤n1+ 5. Hence 2n1≤n2≤n1+ 5 andn1≤5, a contradiction.
We recall a technique frequently used in [13] and then provide a minor modification of [13, Lemma 5.3].
Lemma 4.4. Let Gbe a finite abelian group and letS∈ F(G)be a zero-sum free sequence.
If Q
g∈supp(S)(1 +vg(S))>|G|, then there is anA∈ A(G)with |A| ≥3 such that (−A)A|(−S)S.
Proof. We observe that
|{T ∈ F(G)|T is a subsequence ofS}|= Y
g∈supp(S)
(1 +vg(S)). Thus, ifQ
g∈supp(S)(1 +vg(S))>|G|, then there exist distinct sequencesT1′, T2′ ∈ F(G) such thatT1′ |S, T2′ | S, andσ(T1′) =σ(T2′). We set T1′ =T T1 and T2′ =T T2 where T = gcd(T1′, T2′) andT1, T2∈ F(G).
Thenσ(T1) =σ(T2) and (−T1)T2is a zero-sum subsequence of (−S)S. LetA∈ A(G) withA|(−T1)T2. Assume to the contrary that|A| = 2. Then A = (−g)g for some g ∈ G. Since S is zero-sum free, we infer (after renumbering if necessary) that (−g)|(−T1) and g | T2, a contradiction to gcd(T1, T2) = 1.
Therefore we obtain that |A| ≥ 3, which implies that |gcd(A,(−g)g)| ≤ 1 for each g ∈ G, and thus
(−A)A|(−S)S.
Lemma 4.5. Let t ∈ N andα, α1, . . . , αt, α′1, . . . , α′t ∈R with α1 ≥ . . . ≥αt≥ 0, α′1 ≥. . . ≥ α′t ≥0, α′i≤αi for each i∈[1, t], and Pt
i=1αi≥α≥Pt
i=1α′i. Then
t
Y
ν=1
(1 +xν) is minimal over all(x1, . . . , xt)∈Rt withα′i≤xi≤αi for eachi∈[1, t]andPt
i=1xi=α, if xi=αi for each i∈[1, s] and xi=α′i for each i∈[s+ 2, t]
wheres∈[0, t]is maximal withPs
i=1αi≤α.
Proof. Since continuous functions attain minima on compact sets, the above function has a minimum at some point (m1, . . . , mt) ∈ Rt. Suppose there are i, j ∈ [1, t] such thati < j and mi < mj. Then α′j ≤α′i ≤mi < mj ≤αj ≤αi, and thus we can exchange mi and mj. Therefore, after renumbering if necessary, we may suppose thatm1≥ · · · ≥mt. Since forx≥y ≥0 andδ >0 we have
(1 +x+δ)(1 +y−δ) = (1 +x)(1 +y)−δ(x−y)−δ2<(1 +x)(1 +y),
it follows that all but at most one of the mi is equal to αi or α′i. It remains to show that there is an s ∈ [1, t] such that mi = αi for i ∈ [1, s] and mi = α′i for each i ∈ [s+ 2, t]. Assume to the contrary that this is not the case. Then there are i, j∈[1, t] withi < j such thatmi < αi and α′j < mj. Using again the just mentioned inequality and thatmi≥mj, we obtain a contradiction to the minimum being
attained at (m1, . . . , mt).
Proof of Proposition 4.1. Assume to the contrary that there is anA∈ B(G) such thatL(A) ={2, n2, n1+ n2−2}. Then there areU, V ∈ A(G) with|U| ≥ |V| such thatA=U V. We set
U =SU′ and V = (−S)V′, where U′, V′ ∈ F(G), andS = gcd(U,−V).
Since 2(n1+n2−2)≤ |A|=|U|+|V| ≤2D(G) = 2(n1+n2−1), there are the following three cases:
(I) |A|= 2(n1+n2−2). Then, a factorization ofAof lengthn1+n2−2 must contain only minimal zero-sum sequences of length 2 and thusU′ =V′ = 1.
(II) |A|= 2(n1+n2−2) + 1. Then, a factorization ofAof lengthn1+n2−2 must contain one minimal zero-sum of length 3 and otherwise only minimal zero-sum sequences of length 2, thus U′ =g1g2
andV′ = (−g1−g2) for someg1, g2∈G.
(III) |A| = 2(n1+n2−1). Then a factorization of A of length n1+n2−2 must contain either one minimal zero-sum subsequence of length 4 and otherwise minimal zero-sum sequences of length 2, or two minimal zero-sum sequences of length 3 and otherwise only minimal zero-sum sequences of length 2. Thus, there are the following two subcases.
– U′=g1g2, V′=h1h2where g1, g2, h1, h2∈Gsuch thatg1g2h1h2∈ A(G).
– U′=g1g2(−h1−h2) andV′=h1h2(−g1−g2) whereg1, g2, h1, h2∈G.
We start with the following two assertions.
A1. LetW ∈ A(G) with|W|<|U|andW|(−S)S. Then|W| ∈ {2, n1}.
A2. LetW1, W2∈ A(G) such thatW1(−W1)W2(−W2)|S(−S). Then{|W1|,|W2|} 6={n1}.
Proof of A1. Suppose |W| > 2. Then (−W)W|(−S)S and we set (−S)S = (−W)W T(−T) with T ∈ F(G) and obtain that
U V = (−W)W T(−T)(U′V′).
Let z be a factorization of U′V′. Then |z| ∈ [0,2]. If T = 1, then U V has a factorization of length 2 +|z| ∈ {2, n2, n1+n2−2} which implies |z|= 0 and hence |W|=|U|, a contradiction. Thus T 6= 1.
Since T(−T) has a factorization of length |T| = |S| − |W|, the above decomposition gives rise to a factorization ofU V of lengthtwhere
3≤t= 2 +|T|+|z|= 2 +|S| − |W|+|z| ∈ {n2, n1+n2−2}. We distinguish four cases.
Suppose thatU′ =V′= 1. Thenz= 1,|z|= 0, and|S|=|U|=n1+n2−2. Thust=n1+n2− |W| ∈ {n2, n1+n2−2}, and the assertion follows.
Suppose thatU′ =g1g2andV′ = (−g1−g2) for someg1, g2∈S. Then|z|= 1 and |S|=n1+n2−3.
Thust=n1+n2− |W| ∈ {n2, n1+n2−2}, and the assertion follows.
Suppose thatU′=g1g2,V′ =h1h2 whereg1, g2, h1, h2 ∈Gsuch thatg1g2h1h2∈ A(G). Then |z|= 1 and|S|=n1+n2−3. Thust=n1+n2− |W| ∈ {n2, n1+n2−2}, and the assertion follows.
Suppose thatU′=g1g2(−h1−h2) andV′ =h1h2(−g1−g2) whereg1, g2, h1, h2∈G. Then|z|= 2 and
|S|=n1+n2−4. Thust=n1+n2− |W| ∈ {n2, n1+n2−2}, and the assertion follows. (A1) Proof of A2. Assume to the contrary that |W1|= |W2| =n1. Then there are W5, . . . , Wk ∈ A(G) such that
U V =W1(−W1)W2(−W2)W5·. . .·Wk, wherek∈L(U V) ={2, n2, n1+n2−2} and hencek=n2. Since
|W5·. . .·Wk|=|U V| −4n1≤2(n1+n2−1)−4n1= 2(n2−n1−1), it follows that
k−4≤maxL(W5·. . .·Wk)≤ |W5·. . .·Wk|/2≤n2−n1−1< n2−4,
a contradiction. (A2)
We distinguish two cases depending on the size ofh(S).
CASE 1: h(S)≥n2/2.
We setS=gvS′ whereg∈G,v=h(S), andS′∈ F(G). Then
U1= (−g)n2−vS′U′ ∈ B(G), V1=gn2−v(−S′)V′ ∈ B(G). Clearly, we have
(U′)−1U1= (−g)n2−vS′=− (V′)−1V1 .
We will often use that if some W ∈ A(G) divides (U′)−1U1, then (−W) divides (V′)−1V1 and hence (−W)W|(−S)S. Now we choose factorizationsx1∈Z(U1) andy1∈Z(V1). Note that |x1| ≤n2−v and
|y1| ≤n2−v as each minimal zero-sum sequence in x1 andy1 contains (−g) and g, respectively. Then U V =U1V1 (−g)g2v−n2
has a factorization of lengtht where
2 + (2v−n2)≤t=|x1|+|y1|+ (2v−n2)≤2(n2−v) + (2v−n2) =n2.
Assume to the contrary thatt = 2. Then v =n2/2 and both, U =gn2/2S′U′ andU′ = (−g)n2/2S′U′, are minimal zero-sum sequences, a contradiction, as SU′ ∈ A B/ hgi(G)
as its length is greater than n1 = Dhgi(G) = D(G/hgi). Thus t = n2, |x1| =|y1| =n2−v, and hence L(U1) = L(V1) = {n2−v}.
If W ∈ A(G) with |W|= 2 and W|U1, then W = (−g)g. Similarly, if W′ ∈ A(G) with |W′| = 2 and W′|V1, thenW = (−g)g. By definition of v, not bothU1 and V1 are divisible by an atom of length 2.
Now we distinguish four cases depending on the form ofU′ andV′, which we determined above.
CASE 1.1: U′=V′= 1.
Then V1 = −U1, say V1 = W1·. . . ·Wn2−v. Since none of the Wi has length 2, it follows that
|W1|=. . .=|Wn2−v|=n1 and hence
2(n2+n1−2) = 2|V|= 2(2v−n2) + 2|V1|= 2(2v−n2) + 2n1(n2−v), which implies thatv=n2−1. Consequently|S|=n1−1 andS∈ A Bhgi(G)
. This implies that (use an elementary direct argument or [11, Theorem 5.1.8]),
S= (2e1+a1e2)
n1−1
Y
ν=2
(e1+aνe2),
where (e1, e2) is a basis ofG. Letr∈[0, n2−1] such thatr≡ −a1+a2+a3 modn2. Then
W1= (2e1+a1e2)(−e1−a2e2)(−e1−a3e2)er2 and W2= (2e1+a1e2)(−e1−a2e2)(−e1−a3e2)(−e2)n2−r are minimal zero-sum sequences dividing (−V)V. Since|W1|= 3 +r, |W2|= 3 +n2−r, and|W1W2|= n2+ 6>2n1, at least one of them does not have length n1, a contradiction.
CASE 1.2: U′=g1g2 andV′= (−g1−g2) for someg1, g2∈G.
We set
x1=X1·. . .·Xn2−v and y1=Y1·. . .·Yn2−v
where allXi, Yj ∈ A(G),g1g2|X1X2(or eveng1g2|X1), and (−g1−g2)|Y1. We distinguish three subcases.
CASE 1.2.1: v=n2−1.
By Corollary 4.3, with all notations as introduced there, we getU = en22−1Qn1
ν=1(e1+xνe2). Thus g=e2 andU′|Qn1
ν=1(e1+xνe2), whence after renumbering if necessary we havegi=xie1+e2 for each i∈[1,2]. Therefore we have
V = −2e1−(x1+x2)e2
(−e2)n2−1
n1
Y
ν=3
(−e1−xνe2) = (−g1−g2)(−S).
Assume to the contrary that there are i, j ∈ [3, n1] distinct with xi 6= xj. If q ∈ [1, n2 −1] with q≡ −(xi−xj) modn2, then
W1′ = (e1+xie2)(−e1−xje2)eq2 and W2′ = (e1+xie2)(−e1−xje2)(−e2)n2−q
